Question

Among the following species, the $H-X-H$ angle $(X=B,NorP)$ follows the order:

• A. $P{H}_3<N{H}_3<N{H}_4^{+}<B{F}_3$
• B. $N{H}_3<P{H}_3<N{H}_4^{+}<B{F}_3$
• C. $B{F}_3<P{H}_3<N{H}_4^{+}<B{F}_3$
• D. $B{F}_3<N{H}_4^{+}<N{H}_4^{+}<B{F}_3$

Answer 1

Answer: (A)

$P{H}_3<N{H}_3<N{H}_4^{+}<B{F}_3$

$B{F}_3$ has $H-X-H$ bond angle of $120°$.

$N{H}_4^{+}$ has $s{p}^3$ hybridization due to which the bond angle $=109.28°$. $N{H}_3$ has $s{p}^3$ hybridization but due to lone pair$-$ bond pair repulsion, the $H-X-H$ bond angle decreases to $107°$. In $P{H}_3$, the $X-H$ bonds are longer due to bigger size of $P$, due to which there is more lone pair$-$ bond pair repulsion. Hence, the $H-X-H$ angle is $91°$.

$\therefore$ Order of the bond angle is : $P{H}_3<N{H}_3<\stackrel{⨁}{N}{H}_4<B{F}_3$